# I Algebraic Extensions - Dummit and Foote, Propn 11 and 12 ...

1. May 16, 2017

### Math Amateur

I am reading Dummit and Foote, Chapter 13 - Field Theory.

I am currently studying Section 13.2 : Algebraic Extensions

I need some help with an aspect of Propositions 11 and 12 ... ...

Propositions 11 and 12 read as follows:

Now Proposition 11 states that the degree of $F( \alpha )$ over $F$ is equal to the degree of the minimum polynomial ... ... that is

$[ F( \alpha ) \ : \ F ] = \text{ deg } m_\alpha (x) = \text{ deg } \alpha$

... ... BUT ... ...

... ... Proposition 12 states that ... "if $\alpha$ is an element of an extension of degree $n$ over $F$, then $\alpha$ satisfies a polynomial of degree at most $n$ over $F$ ... ... "

Doesn't Proposition 11 guarantee that the polynomial (the minimum polynomial) is actually of degree equal to $n$???

Can someone please explain in simple terms how these statements are consistent?

Help will be appreciated ...

Peter

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2. May 16, 2017

### Staff: Mentor

1. In order to prove ($\alpha$ algebraic $\Rightarrow \, F(\alpha) /F$ finite), we don't care exact degrees.
2. The authors mentioned, that $\deg [F(\alpha):F] = \deg m_\alpha(x)$.
3. In the direction of the proof which you quoted, all we have is that $\alpha$ is algebraic over $F$. This means it satisfies some polynomial equation of degree $n$. This polynomial doesn't need to be minimal, irreducible nor has $\alpha$ to be outside of $F$. That it is of finite degree is all that counts. We simply don't bother more than that.

3. May 16, 2017

### andrewkirk

This one is easier than your usual challenges!

The trouble arises because the $\alpha$ in the second sentence is not necessarily the same as the one in the first sentence. It might have been clearer if they'd used $\beta$ instead of $\alpha$ in the second and subsequent sentences.

Let the extension of interest be $F(\alpha) / F$ where $\alpha$ is the root of an irreducible quadratic in $F[x]$.
Consider a $\beta$ that is in $F$. That $\beta$ is also in the extension. But it is a root of the degree-1 $F[x]$ polynomial $x-\beta$.

On the other hand, $\alpha$ is also in the extension, and the minimal polynomial for that has degree two.

So some elements of the extension (the ones that are not already in $F$) have degree two, and some (those that are in $F$) have degree one.

4. May 17, 2017

### Math Amateur

Thanks fresh_42 and Andrew ... appreciate your help ...

Still thinking over what you have said ...

Peter